Question: Factor completely. $4 -12x +9x^2=$
Solution: Both $4$ and $9x^2$ are perfect squares, since $4=({2})^2$ and $9x^2=({3x})^2$. Additionally, $12x$ is twice the product of the roots of $4$ and $9x^2$, since $12x=2({2})({3x})$. $4-12x+9x^2 = ({2})^2 - 2({2})({3x})+({3x})^2$ So we can use the square of a difference pattern to factor: ${a}^2 - 2( a)( b)+ {b}^2 =({a} - {b})^2$ In this case, ${a}={2}$ and ${b}={3x}$ : $ ({2})^2 - 2({2})({3x})+({3x})^2 =({2} - {3x})^2$ In conclusion, $4 -12x +9x^2=(2 -3x)^2$ Remember that you can always check your factorization by expanding it.